NIELIT 2017 DEC Scientist B - Section B: 25

Question

Consider the relational schema $R(A B C D)$ with following $FD$ set $F=\{A \to CE, B \to D, AE \to D\}$. Identify the highest normal form satisfied by the relation $R$.

• A. $2$NF
• B. BCNF
• C. $3$NF
• D. $1$NF

Comments

When the relational schema is $R(A, B, C, D)$ then what is $E$ in FD $A \rightarrow CE$ and $AE\rightarrow D$? Are attributes not present in schema allowed in FDs?

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E is not present in relation but there in fd? is this possible. Can anyone please clarify this.

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yes just ignore E in the fd set thats all

Answer 1

F = { A→CE, B→D, AE→D }

Here candidate key is: AB

The functional dependency B→D violates 2NF requirement. It is a partial dependency, D is partially dependent on B, where B is a proper subset of a candidate key.

Hence, it is in (D) 1NF

Answer 2

FD set F={ A→CE, B→D, AE→D }
AB+={A,B,C,D,E} it clearly shows that right side, there is no B. So, definitely candidate key require B.
→ The functional dependency B→D violates 2NF requirement. It is a partial dependency, D is partially dependent on B, where B is a proper subset of a candidate key.

Answer 3

Here, $AB$ is the candidate key as $AB^{+}=ABCDE$

Here $B\rightarrow D$ is a partial dependency as a prime attribute $B$ is determining non-prime attribute $D$, which is not allowed in $2$NF.

$\therefore$ The highest normal form satisfied by the relation is $1$NF

Option D is correct.

Comments

"a prime attribute B is determining non-prime attribute D, which is not allowed in 2NF."

This statement is not correct

Answer 4

ANS : D 1ST NORMAL FORM

F={A→CE,B→D,AE→D}

CANDIDATE KEY : AB → ABCDE

IT IS PARTIALLY DEPENDENT

B→D VIOLET THE CONDITION OF 2ND NORMAL FORM

SO ,GIVEN RELATION IS ONLY IN 1ST NORMAL FORM .